= \(\frac{1.3-1}{1.3}+\frac{3.5-1}{3.5}+...+\frac{17.19-1}{17.19}=1-\frac{1}{1.3}+1-\frac{1}{3.5}+...+1-\frac{1}{17.19}\)

= \(\left(1+1+...+1\right)-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{17.19}\right)\)

= \(9-\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{17.19}\right)=9-\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{17}-\frac{1}{19}\right)\)

= \(9-\frac{1}{2}.\left(1-\frac{1}{19}\right)=9-\frac{1}{2}.\frac{18}{19}=9-\frac{9}{19}=\frac{162}{19}\)

S = \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+\frac{5}{7.9}+.......+\frac{5}{17.19}\)

S : 5 = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{17.19}\) 

S : 5 = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}+.......+\frac{1}{17}-\frac{1}{19}\)

=> S : 5 = 1 - \(\frac{1}{19}=\frac{19}{19}-\frac{1}{19}=\frac{18}{19}\)

=> S = \(\frac{18}{19}x5=\frac{90}{19}\)

=5/19

sai bet thang ngu nhu cho

D= 1 x 99 x 36 cặp = 3564

\(P=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2021.2023}\)

\(2P=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{3}{5.7}+...+\dfrac{2}{2021.2023}\)

\(2P=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2023}\)

\(2P=\dfrac{1}{1}-\dfrac{1}{2023}\)

\(P=\dfrac{2022}{2023}:2\)

\(P=\dfrac{1011}{2023}\)

\(=>P=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2023}\)

\(P=1-\dfrac{1}{2023}=\dfrac{2023}{2023}-\dfrac{1}{2023}=\dfrac{2022}{2023}\)

\(x.P=\dfrac{2022}{2023}=>x=P:\dfrac{2022}{2023}=\dfrac{2022}{2023}:\dfrac{2022}{2023}=1\)

\(\frac{4}{1.3}+\frac{16}{3.5}+\frac{36}{5.7}+...+\frac{324}{17.19}\)

\(=\frac{2^2}{1.3}+\frac{4^2}{3.5}+\frac{6^2}{5.7}+...+\frac{18^2}{17.19}\)

\(=2\left(\frac{2.1^2}{1.3}+\frac{2.2^2}{3.5}+\frac{2.3^2}{5.7}+...+\frac{2.9^2}{17.19}\right)\)

\(=2\left(\frac{1}{1}-\frac{1}{3}+\frac{2^2}{3}-\frac{2^2}{5}+\frac{3^2}{5}-\frac{3^2}{7}+...+\frac{9^2}{17}-\frac{9^2}{19}\right)\)

\(=2\left[1+\frac{2^2-1}{3}+\frac{3^2-2^2}{5}+...+\frac{9^2-8^2}{17}-\frac{9^2}{19}\right]\)

\(=2\left(1+1+1....+1-\frac{81}{19}\right)=2\left(9-\frac{81}{19}\right)=\frac{180}{19}\)